Test Index

Sequence and Series

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Question : 48 of 86

Marks: +1, -0

Which term of the A.P.

\; \frac{3}{\sqrt{7}},\; \frac{4}{\sqrt{7}},\; \frac{5}{\sqrt{7}}, \dots

\; \frac{17}{\sqrt{7}} ?

13
14
15
16

Solution:

Concept:

n

th term of an A.P. is

a + (n-1)d

.

Explanation:

Here, first term

a = \frac{3}{\sqrt{7}}

and common difference

d = \frac{4}{\sqrt{7}} - \frac{3}{\sqrt{7}} = \frac{1}{\sqrt{7}}

.

Let the

n

th term be

\frac{17}{\sqrt{7}}

.

Then,

a + (n-1)d = \frac{3}{\sqrt{7}} + (n-1)\frac{1}{\sqrt{7}} = \frac{n+2}{\sqrt{7}}

.

Set

\frac{n+2}{\sqrt{7}} = \frac{17}{\sqrt{7}} \Rightarrow n+2 = 17 \Rightarrow n = 15

.

Thus, the 15th term is

\frac{17}{\sqrt{7}}

.

Answer:

15

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